# Properties of Parallelograms Practice with Our Examples

Recent questions in Properties of parallelograms
Oswaldo Riley 2023-03-25

## State whether the statements are True or False. All rectangles are parallelograms.

warnest1il 2023-03-13

## ABCD is a parallelogram. Which of the following is not true about ABCD? AB=CD AD=CB AD||CD AB||CD

Jaslyn Gordon 2023-02-25

## Which of the following is not a parallelogram? A)Square B)Rectangle C)Rhombus D)Trapezium

otium9ag 2023-02-06

## Which of the following is not a necessary condition for a quadrilateral ABCD to be a parallelogram:A)AB = CDB)AB // CDC)$\angle A={50}^{\circ },\mathrm{\angle }C={50}^{\circ },\mathrm{\angle }B={130}^{\circ }$D)AB⊥AD

markezop78 2023-02-01

## How can magnets lose their properties? A. If they are heated B. If they are hammered C. If they are dropped from height D. All of these

Jovani Mayer 2023-01-27

## All rectangles are parallelograms. A)True B)False

Charlee Sweeney 2023-01-20

## What is the volume of a block of ice measuring 3 meters long, 1.5 meters wide, and 2 meters high?

Arnav Colon 2023-01-08

## State whether True or False.(1) All rectangles are squares.(2) All rhombuses are parallelograms.(3) All squares are rhombuses and also rectangles.(4) All squares are not parallelograms.(5) All kites are rhombuses.(6) All rhombuses are kites.(7) All parallelograms are trapeziums.(8) All squares are trapeziums.

Aron Pierce 2022-12-22

## PQ and RS are two equal and parallel line segments. Any point M not lying on PQ or RS is joined to Q and S and lines through P parallel to QM and through R parallel to SM meet at N. Prove that line segments MN and PQ are equal and parallel to each other.

Fletcher Ramirez 2022-11-29

## Which statements about a parallelogram are accurate? A Diagonals cut each other in half. B The parallelogram is divided into two congruent triangles by its diagonals. C In a parallelogram, the following angles are supplementary. D A parallelogram's opposing angles are always supplementary.

Sonia Elliott 2022-10-20

## I don't understand what this theorem for a characterization of the determinant actually means, and what use it has. Could you please explain it?Let D be a function mapping from the set of square matrices M with n rows/columns over the field F, to a field F. Also let D be a function that is multilinear and alternating in columns. ThenD(A)=D(I)detA

elisegayezm 2022-09-30

## Let C[0,1] be the space of continuous functions $\left[0,1\right]\to \mathbb{R}$ endowed with the norm $||x|{|}_{\mathrm{\infty }}={\mathrm{m}\mathrm{a}\mathrm{x}}_{t\in \left[0,1\right]}|x\left(t\right)|$. It is easy to verify that this norm is not induced by any inner product (really the parallelogram law fails for x(t)=t and y(t)=1). Well, how to understand that this norm is not equivalent to anyone induced by an inner product? So, the norms induced by inner products should have some special properties...

Liam Potter 2022-09-17

## Prove that there is an inner product on ${\mathbb{R}}^{2}$, such that the associated norm is given by:$\parallel \left(x,y\right)\parallel =\left(|x{|}^{p}+|y{|}^{p}{\right)}^{\frac{1}{p}}$where p>0 only if p=2So far what I have tried to do is assume there exists such an inner product for some aribtrary p, and then show that some property that holds for all inner products (e.g. the Cauchy-Schwarz inequality, triangle inequality, parallelogram equality, does not hold for the inner product with the said associated norm unless p=2.First I tried the parallelogram equality and ended up with:$\left(|{x}_{1}+{x}_{2}{|}^{p}+|{y}_{1}+{y}_{2}{|}^{p}{\right)}^{\frac{1}{p}}=\left(|{x}_{1}|+|{y}_{1}|{\right)}^{\frac{1}{p}}+\left(|{x}_{2}{|}^{p}+|{y}_{2}{|}^{p}{\right)}^{\frac{1}{p}}$but I don't know how to show that this equality only holds for p=2 (although I'm pretty sure it does because I tried plugging in random values for $p\ne 2$).Since for an inner product, the parallelogram equality must hold,$⟨u,v⟩=\frac{1}{2}\left(\parallel u+v{\parallel }^{2}+\parallel u-v{\parallel }^{2}\right)$must also hold.Using this definition of the inner product, I also tried to show a contradiction by showing that if $p\ne 2$, the Cauchy-Schwarz Inequality didn't hold. However I think that's a dead end.

Elementary geometryOpen question
sarahkobearab4 2022-08-18

## I came across a property which I am not sure if true in general. Suppose you have a parallelogram whose vertices are ${v}_{1},{v}_{2},{v}_{3},{v}_{4}\in {\mathbb{R}}^{2}$. Let's say that the side $\left[{v}_{1},{v}_{4}\right]$ is parallel to $\left[{v}_{2},{v}_{3}\right]$. Is it true that$||{v}_{1}-{v}_{4}|{|}^{2}<||{v}_{1}-{v}_{3}||\cdot ||{v}_{2}-{v}_{4}||,$that is, the square of one side is less than the product of the diagonals? I could also be just missing an obvious counterexample, but so far I can't prove it either.

Elementary geometryOpen question
Ebone6v 2022-08-17

## According to wikipedia article on polarisation identity, in a normed space $\left(V,||.||$, if the parallelogram law holds, then there is an inner product on V such that $||x|{|}^{2}=⟨x,x⟩$ for all $x\in V$. However, from what I understand, we can prove polarisation identity this way:$||x+y|{|}^{2}-||x-y|{|}^{2}=⟨x+y,x+y⟩-⟨x-y,x-y⟩=...=4⟨x,y⟩$However, from this proof, I don't see why we need parallelogram law. On top of that, shouldn't it be the other way round, such that if $||x|{|}^{2}=⟨x,x⟩$, then the parallelogram law holds? Can someone explain to me please?

Crancichhb 2022-08-12

## It can be proved easily by contradiction thatif a,b,c,d are positive numbers, then$\frac{a+b}{c+d}\ge min\left\{\frac{a}{c},\frac{b}{d}\right\}.$I am not looking for a proof but rather for1) a reference or book which contain this and similar inequalities;2) information whether this inequality can be sharpened.

Matonya 2022-08-10